Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**Mathfun****Guest**

Let "a" and "b" be real numbers. We have 2 functions:

f(x) = ax + b|x| and g(x) = ax - b|x|

Show that if: f(f(x)) = x for every real x, then:

g(g(x)) = x for every real x

Pleas help 'cause I haven't got any idea how to solve it.

**mathfun****Guest**

No one can help?? PLEASE It's veary important for me I have to present it tomorrow in class!!

**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

And with that post, kylekatarn becomes a ** power member**. Well done to you.

Back on topic, I tried it and got f(f(x)) = a (ax + b|x|) + b|(ax + b|x|)|.

For g(g(x)) to equal x when f(f(x)) is equal to x, then f(f(x) would have to be equal to g(g(x) and so b would have to be 0.

So now you have to prove that a (ax + b|x|) + b|(ax + b|x|)| ≠ x for all values of x when b ≠ 0. That's the tricky bit.

Why did the vector cross the road?

It wanted to be normal.

Offline

**kylekatarn****Member**- Registered: 2005-07-24
- Posts: 445

f(f(x)) = x => g(g(x)) = x

g(g(x))=f(f(x))

f(x)=ax+b|x|

g(x)=ax-b|x|

----------------------------------

f(f(x))=a.f(x)+b|f(x)|

g(g(x))=a.g(x)-b|g(x)|

----------------------------------

a.f(x)+b|f(x)|=a.g(x)-b|g(x)|

:: k=b/a ::

f(x)+k|f(x)|=g(x)-k|g(x)|

f(x)-g(x)=-k|g(x)|-k|f(x)|

g(x)-f(x)=k|g(x)|+k|f(x)|

g(x)-f(x)=k(|g(x)|+|f(x)|)

g(x)-f(x)=k(|g(x)|+|f(x)|)

g(x)-f(x) = ax - b|x| - (ax + b|x|) = ax - b|x| - ax - b|x|) = -2b|x|

-2b|x|=k(|g(x)|+|f(x)|)

-2b(1/k)|x|=|g(x)|+|f(x)|

:: 1/k=a/b ::

-2b(a/b)|x|=|g(x)|+|f(x)|

-2a|x|=|g(x)|+|f(x)|

That's as far as I can gow for now.

You can try other things like: triangular inequality, proofing by expanding the module or proof by contradiction.

Best luck for you now.

*Last edited by kylekatarn (2005-11-03 05:42:54)*

Offline

**kylekatarn****Member**- Registered: 2005-07-24
- Posts: 445

mathsyperson wrote:

And with that post, kylekatarn becomes a

. Well done to you.power member

Btw, thanks! : )

Offline

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,664

Congrats kylekatarn, and thank you on behalf of everyone you have helped to get you there! (That was a curious sentence wasn't it?)

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

Offline

**kylekatarn****Member**- Registered: 2005-07-24
- Posts: 445

...indeed:)

Again, thank you all:)

Offline

**Mathfun****Guest**

I still can't fix it..... ;(

**kylekatarn****Member**- Registered: 2005-07-24
- Posts: 445

I finnaly got it. I don't know if you still want the solution but here it is, anyway:

----------------------------------------------------------------------------------------------------------

We want to show that if f( f(x) ) = x then g( g(x) ) = x, for all x∊ℝ

H1 (Hipothesis):

∀x∊ℝ f( f(x) ) = x

T1 (Thesis):

∀x∊ℝ g( g(x) ) = x

This is an implication where

H1 ⇒ T1

∀x∊ℝ [ f(f(x)) = x ⇒ g(g(x)) = x ]

We will determine the conditions for the Hipothesis to be true and then do the same for the Thesis. The we find the conditions are the same for both Hipothesis and Thesis. This allows us to conclude that the inicial statement is more than an implication, its something "logically stronger" - an equivalence H1⇔T1

----------------------------------------------------------------------------------------------------------

L1 (Lemma 1):

∀x≥0 ⇒ |x| = x

∀x<0 ⇒ |x| = -x

(the case x=0 was included in the first implication, but it could be on the second one, too)

Proof: Its obvious, considering the definition of absolute value.

----------------------------------------------------------------------------------------------------------

Proof of H1:

∀x∊ℝ f( f(x) ) = x

H1.a) Case x≥0

f(x) = ax+b|x|

f(x) = ax+bx //by L1

f(x) = (a+b)x

f( f(x) ) = f( (a+b)x ) = (a+b)²x

f( f(x) ) = x

(a+b)²x = x ⇔ (a+b)² = 1

Conclusion from H1.a)

∀x≥0 [ f( f(x) ) = x ] ⇔ (a+b)² = 1

H1.b) Case x<0

f(x) = ax+b|x|

f(x) = ax+b(-x) //by L1

f(x) = ax-bx

f(x) = (a-b)x

f( f(x) ) = f( (a-b)x ) = (a-b)²x

f( f(x) ) = x

(a-b)²x = x ⇔ (a-b)² = 1

Conclusion from H1.b)

∀x<0 [ f( f(x) ) = x ] ⇔ (a-b)² = 1

Conclusion from H1.a) and H1.b)

∀x≥0 [ f( f(x) ) = x ] ⇔ (a+b)² = 1

and

∀x<0 [ f( f(x) ) = x ] ⇔ (a-b)² = 1

therefore

∀x∊ℝ [ f( f(x) ) = x ] ⇔ [ (a+b)² = (a-b)² =1 ]

----------------------------------------------------------------------------------------------------------

Proof of T1:

∀x∊ℝ g( g(x) ) = x

T1.a) Case x≥0

g(x) = gx-b|x|

g(x) = ax-bx //by L1

g(x) = (a-b)x

g(g(x)) = g((a-b)x) = (a-b)²x

g(g(x)) = x

(a-b)²x = x ⇔ (a-b)² = 1

Conclusion from T1.a)

∀x≥0 [ g( g(x) ) = x ] ⇔ (a-b)² = 1

H1.b) Case x<0

g(x) = ax-b|x|

g(x) = ax-b(-x) //by L1

g(x) = ax+bx

g(x) = (a+b)x

g( g(x) ) = f( (a+b)x ) = (a+b)²x

g( g(x) ) = x

(a+b)²x = x ⇔ (a+b)² = 1

Conclusion from T1.b)

∀x<0 [ g( g(x) ) = x ] ⇔ (a+b)² = 1

Conclusion from T1.a) and T1.b)

∀x≥0 [ g( g(x) ) = x ] ⇔ (a-b)² = 1

and

∀x<0 [ g( g(x) ) = x ] ⇔ (a+b)² = 1

therefore

∀x∊ℝ [ g( g(x) ) = x ] ⇔ [ (a-b)² = (a+b)² =1 ]

----------------------------------------------------------------------------------------------------------

Conclusions from H1 and T1

∀x∊ℝ [ f( f(x) ) = x ] ⇔ [ (a+b)² = (a-b)² =1 ]

and

∀x∊ℝ [ g( g(x) ) = x ] ⇔ [ (a-b)² = (a+b)² =1 ]

therefore

∀x∊ℝ [ f( f(x) ) = x ] ⇔ ∀x∊ℝ [ g( g(x) ) = x ] //by the transitivity property of '⇔'

∀x∊ℝ [ f( f(x) ) = x ⇔ g( g(x) ) = x ]

so we can say that

H1⇔ T1 ∴ H1⇒ T1 //this is what we wanted to prove

and more..

T1⇒ H1

*Last edited by kylekatarn (2005-11-05 04:46:17)*

Offline

**Mathfun****Guest**

Big BIg BIg BIg BIg thanks!!!!!!!!!

**kylekatarn****Member**- Registered: 2005-07-24
- Posts: 445

no problem : )

Offline

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,664

Most impressive, kylekatarn.

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

Offline

Pages: **1**